Question 1013006

{{{ x^2 -(k-2)x-12=0}}} 

to have roots that are equal but opposite in sign, we can write our equation like this:

{{{ x^2 -(k-2)x-12=0}}}
{{{ (-x)^2 -(k-2)(-x)-12=0}}}

And since we started with {{{ x^2 -(k-2)x-12=0}}}

we can set the left sides equal:

{{{ x^2 -(k-2)x-12=  (-x)^2 -(k-2)(-x)-12 }}}....solve for {{{k}}}

{{{ x^2 -kx+2x-12= x^2 +kx-2x-12 }}}

{{{ cross(x^2 )-k*cross(x)+2cross(x)-cross(12)= cross(x^2) +kx-2cross(x)-cross(12) }}}

{{{ -k+2= k-2 }}}

{{{ 2+2= k+k }}}

{{{ 4= 2k }}}

{{{k=2}}}


check:

{{{ x^2 -(k-2)x-12=0}}} 

{{{ x^2 -(2-2)x-12=0}}} 

{{{ x^2 -(0)x-12=0}}} 

{{{ x^2-12=0}}} 

{{{ x^2=12}}} 

{{{ x=sqrt(12)}}} 

{{{ x=sqrt(4*3)}}} 

{{{ x=2sqrt(3)}}} 

solutions: {{{ x=2sqrt(3)}}} and {{{ x=-2sqrt(3)}}}